26. The Laplace transform of i(t) is given by i(s) = 2/s(1 + s). as t the value of i(t) tends to

(a) 0

(b) 1

(c) 2

(d) 0

**The Fourier series expansion of a real periodic signal with fundamental frequency f**_{o}is given by g_{p}(t) = c_{0}e^{j2nft}. it is given that c_{3}= 3 + j5.

(a) 5 + j3

(b) -3 – j5

(c) -5 + j3

(d) 3 – j5

- Let x(t) be the input to a linear, time-invariant system. the required output is 4x(t – 2). the transfer function of the system should be

(a) 4e^{j4t}

(b) 2e^{-j8t}

(c) 4e^{-j4t}

(d) 2e^{j8t}

- A sequence x(n) with the z-transform x(z) = z
^{4}+ z^{2}– 2z + 2 – 3z^{-4}is applied as an input to a linear, time-invariant sysyem with the impulse response h(n) = 2(n – 3), where (n) = {1, n = 0 0, otherwise

the output at n = 4 is

(a) -6

(b) zero

(c) 2

(d) -4

- Convolution of x(t + 5) with impulse function (t – 7) is equal to

(a) x(t – 12)

(b) x(t + 12)

(c) x(t – 2)

(d) x(t + 2)

- Which of the following cannot be the fourier series expansion of a periodic signal?

(a) x(t) = 2 cos t + 3 cos 3t

(b) x(t) = 2 cos t + 7 cos t

(c) x(t) = cos t + 0.5

(d) x(t) = 2 cos 1.5t + sin 3.5 t

- The fourier transform f(e
^{-t}u(t) is equal to 1/1 + j2f. therefore, f{1/1 – j2t} is

(a) e^{f} u(f)

(b) e^{-f} u(f)

(c) e^{f} u(-f)

(d) e^{-f} u(-f)

- A linear phase channel with phase delay t
_{p}and group delay t_{g}must have

(a) t_{p} = t_{g} = constant

(b) t_{p} f and t_{g} f

(c) t_{p} = constanf and t_{g} f

(d) t_{p} f and t_{g} = constant

- Conider a sampled signal :

y(t) = 5 x 10^{-6} x(t) (t – nt_{s}),

where, x(t) = 10 cos (8 x 10^{3}) t

and t_{s} = 100 us. when y(t) is passed through an ideal low-pass filter with a cut-off frequency of 5 khz, the output of the filter is

(a) 5 x 10^{-6} cos (8 x 10^{3}) t

(b) 5 x 10^{-5} cos (8 x 10^{3}) t

(c) 5 x 10^{-1} cos (8 x 10^{3}) t

(d) 10 cos (8 x 10^{3}) t

- The transfer function of a system is given by h(s) = 1/s
^{2}(s – 2) . the impulse response of the system is (* denotes convolution, and u(t) is unit step function)

(a) (t^{2} * e^{-2t}) u(t)

(b) (t * e^{2t}) u(t)

(c) te^{-2t}) u(t)

(d) (te^{-2t}) u(t)

- The region of convergence of the z-transform of a unit step function is

(a) |z|>1

(b) |z|<1

(c) (real part of z)> 0

(d) (real part of z) < 0

- Let (t) denotes the delta function. the value of the integral (t) cos (3t/2) dt is

(a) 1

(b) -1

(c) 0

(d) 2

- A band limited signal is sampled at the nyquist rate. the signal can be recovered by passing the samples through

(a) an R-C filter

(b) an envelope detector

(c) a PLL

(d) an ideal low-pass filter with appropriate bandwidth

- A linear time-invariant system has an impulse response e
^{2t}, t > 0. if the initial conditions are zero and the input is e^{3t}, the output for t > 0 is

(a) e^{3t} – e^{2t}

(b) e^{5t}

(c) e^{3t} + e^{3t}

(d) none of these

- Given that l[f(t)] = s + 2/s
^{2}+ 1, l[g(t)] = s^{2}+ 1/(s + 3) (s + 2). h(t) = f(r) g(t) d. l[h(t)] is

(a) s^{2 }+ 1/s + 3

(b) 1/s + 3

(c) s^{2} + 1/(s + 3) (s + 2) + s + 2/s^{2} + 1

(d) none of these

- The fourier transform of the signal x(t) = e
^{-3t2}is of the following form, where A and B are constants

(a) Ae^{-5(t)}

(b) Ae^{-bf2}

(c) A + B |F|^{2}

(d) Ae^{-bf}

- A system with an input x(t) and output y(t) is described by the relation y(t) = tx(t). this system is

(a) linear and time-invariant

(b) linear and time-variant

(c) non-linear and time-invariant

(d) non-linear and time-variant

- The n-point discrete fourier transform of a signal x[n] is x
_{dft}[k]. if DFT of another discrete sequence y[n] is given as y_{dft}[k] = x*_{dft}[k], then which of the following is true?

(a) y[n] = x* [n]

(b) y[n] = nx* [n]

(c) y[n] = 1/n x* [n]

(d) y[n] = – x* [n]

- Discrete fourier transform of a real sequence {x[n]} is given as

x_{dft}[k] = {0, a, 2 + j, – 1,b,j}

then, a and b are

(a) 2 – j, j

(b) j, -j

(c) -j, 2 – j

(d) j, 2 + j

- The 4-point discrete fourier transform of a discrete time sequence (1,2,1,2) is

(a) {0, -2, 0, 6}

(b) {6, -4j, 6 – 4j}

(c) {-4j , 6 – 4j, 6}

(d) {6, 0, -2, 0}

- Consider a band-pass signal x
_{c}(t) is shown below. the minimum sampling rate required for this signal to prevent loss of information is

(a) 4 kHz

(b) 12 kHz

(c) 20 kHz

(d) 2 kHz

- Two signals x
_{1}(t) are band-limited to 2 kHz and 3kHz respectively, then nyquist rate for the signal x_{1}(t)* x_{2}(t) is

(a) 5 kHz

(b) 6 kHz

(c) 4 kHz

(d) 10 kHz

- Which one is most appropriate dynamic system out of the following?

(a) y(n) = y(n – 1) + y(n + 1)

(b) y(n) = y(n – 1)

(c) y(n) = x(n)

(d) y(n) + y(n – 1) + y(n + 3) = 0

- Which one is causal system?

(a) y(n) = 3x(n) – 2x(n – 1)

(b) y(n) = 3x(n) + 2x(n + 1)

(c) y(n) = 3x(n + 1) + 2x(n – 1)

(d) y(n) = 3x(n + 1) + 2x(n – 1) + x (n)

- For the signal given below

y(t) = 2 sin (2/3 t) + 4sin (1/4 t 4) + 6 sin (1/3 t 5) + 8 sin (1/2 t 7)

the common period of y(t) is given by

(a) 12

(b) 24

(c) 8

(d) 16

- (c)

from final value theorem,

lim I(t) = lim sI (s)

= lim s x 2/s (1 + s)

= 2

- (d)

the coefficients c_{n} and c_{-n} in the fourier series are complex conjugate.

C_{-N} = C_{N}*

|C_{-N} |= |C_{N}|

<C_{-N} = – <C_{N}

C_{3} = 3 + J5

C_{-3} = C_{3}* = 3 – J5

- (c)

transfer function h(s) = FT of output/FT if input

= 4e^{-j20} x(s)/x(s) = 4e^{-j2o}

h(s) = 4e^{-j4f}

- (b)

y(z) = h(z) x(z)

given, x(z) = z^{4} + z^{2} – 2z + 2 – 3z^{-4}

h(n) = [n – 3]

h(z) = z^{-3}

y(z) = z^{-3} [z^{4} + z^{2} – 2z + 2 – 3z^{-4}]

= z + z^{-1} + 2z^{-2} + 2z^{-3} – 3z^{-7}

taking inverse z-transform,

by inspection,

y[n] = [n + 1] + [n – 1] – 2[n – 2] + 2 [n – 3] – 3 [n – 7]

putting n = 4

y[4] = 0

- (c)

from shifting property of impulse function,

(t)* (t – t_{o}) = (t – t_{0})

hence, x(t + 5)* (t – 7) = x(t + 5 – 7)

= x(t – 2)

- (b)

signal x(t) = 2 cos t + 7 cos t

period of 2 cos t = t_{1} = 2/0

= 2 = 2

period of 7 cos t = t_{2} = 2/0

= 2/1 = 2

for signal x(t) to be periodic t_{1}/t_{2} not be rational.

t_{1}/t_{2} = 2/2 = 1 = irrational

periodic waveform is sufficient condition for existence of fourier series. hence, x(t) = 2 cos t + 7 cos t is not a periodic signal hence, fourier series expansion is not possible.

- (c)

from duality property of fourier transform.

x(t) – x(f)

x(t) – x(-f)

f{1/1 + j2t} = e^{f} u(-f)

- (a)

let V_{I} = x(t) cos t is applied in linear phase channel.

where, T_{g} = group delay

T_{p} = phase delay

in liear phase channel, transmission is distrtionless.

hence, T_{P} = T_{G} = constant

- (c)

given that sampling interval t_{s} = 100 us

sampling frequency f_{s} = 1/t_{s} = 10 kHz

bandwidth of LPF = f_{c} = 5 kHz

f_{s} = 2 B

hence, sampled output not overlap.

filter output = y(t)/t_{s}

= 5 x 10^{-6} x x(t)/100 x 10^{-6}

given, x(t) = 10 cos (8 x 10^{3}) t

output = 5 x 10^{-6} x 10 cos (8 x 10^{3}) t/100 x 10^{-6}

= 5 x 10^{-1} cos (8 x 10^{3}) t

- (b)

transfer function h(s) = 1/s^{2}(s – 2)

lmpulse response h(t) = l^{-1} [h(s)]

= l^{-1} [1/s^{2}(s – 2)]

we know that

l^{-1} [f_{1} (t). f_{2}(t)] = l^{-1} [f_{1}(t)* l^{-1}[f_{2}(t)]

hence, h(t) = l^{-1} [1/s^{2}]* l^{-1}|1/s – 2]

= t u(t)* e^{2t} u(t)

= (t^{*} e^{2t}) u(t)

- (a)

x[n] = u[n]

x[z] = x[k] z^{-k}

u[k] z^{-k} = z^{-k}

x[z] = 1/1 – z^{-1}

for x[z] to converge,

|z^{-1}|< 1

|z|> 1

- (a)

from sampling property of unit impulse function,

(t) (t) dt = (0)

(t) cos (3t/2) dt = cos (0) = 1

- (d)

As band limited signal is sampled at nyquist rate. hence, replicated spectra do not overlap and the original sepctrum can be generated by passing 0_{s} (t) through ideal low-pass filter that has cut-off frequency.

(bandwidth f_{c } = f_{s}/2 ; where f_{s} is sampling frequency.

- (a)

y(s) = x(s) h(s)

given, x(t) = e^{3t}, h(t) = e^{2t}

x(s) = 1/s – 3, h(s) = 1/s – 2

y(s) = x(s) h(s) = 1/(s – 3) x 1/(s – 2)

= 1/s – 3 – 1/s – 2

hence, y(t) = e^{3t –}– e^{2t}

- (b)

given, l[f(t)] = s + 2/s^{2} + 1 = f(s)

l[g(t)] = s^{2} + 1/(s + 3)(s + 2) = g(s)

h(t) = f g(t-) d

h(t) = f(t)* g(t)

taking laplace transform,

l[h(t) = l[f(t)* g(t)]

= l[f(t)]. l[g(t)]

= f(s). g(s)

= s + 2/s^{2} + 1 x s^{2} + 1/(s + 3) (s + 2) = 1/s + 3

- (b)

the signal x(t) = e^{-3t2} is an example of gaussian distribution. the fourier transform of gaussian distribution is also caussian distribution.

x(f) = A e^{-bf2}

- (b)

let x(t) = a x_{1} (t) + b x_{2} (t)

then, a y_{1} (t) + b y_{2} (t) = at x_{1} (t) + bt x_{2}(t)

= t x(t)

then system is linear.

if input is delayed by t_{0}, i.e., x(t – t_{0})

then output is

y_{1} (t) = t x (t – t_{o})

but output is delayed by t_{o}.

y_{2}(t) = (t – t_{o} x (t – t_{o})

y_{1} (t) = y_{2}(t)

hence, system is not time-invariant, i.e., time varying.

- (b)

x[n] = x_{dft} [k]

a typial DFT term X_{DFT}[K] has the form Ae^{jo}

let = 2kn /n, the IDFT gives x[n] with terms of the form

1/n Ae^{jo} . e^{j} = 1/n Ae^{j(0+ 0)}

if we conjugate X_{DFT} [k], typical term becomes Ae^{-jo}. its IDFT gives terms of the form Ae^{-jo}e^{-jo} = Ae^{-j(0 + 0)}

this DFT result corresponds to the signal Nx *[n].

- (c)

the DFT of a real signal shows conjugate symmetry as

X_{DFT}[K] = X_{DFT} [N – K]

here, N = 6

a = X_{DFT} [1] = X_{DFT} [6 – 1] = X_{DFT} [5]

a = – j

B = X_{DFT} [4] = X_{DFT}[6 – 4] = X_{DFT}[2]

B = 2 – J

- (d)

let x[n] = (1,2,1,2} is discrete fourier transform of x[n] is given as

X_{DFT} [K] = X[n] e^{-j2nk/n}

k = 0, 1,……….N – 1 and

n = 4

so, X_{DFT} [K] = x[n] e^{-j2nk/2}

x[n] e^{-jkn/2}

k = 0, X_{DFT} [0] = x[n] e^{o}

x[0] + x[1] + x[2] + x[3]

= 1 + 2 + 1 + 2 = 6

k = 1, X_{DFT} [1] = x[n] e^{-jn/2}

= x_{1} [0] e^{0} + x[1] e^{-j/2} + x[2] e^{-j} + x[3] e^{j3/2}

= 1 – 2j – 1 + 2j = 0

k = 2, X_{DFT} [2] = x[n] e^{-jn}

= x[0] e^{0} + x[1] e^{-j} + x[2] e^{-j2} + x[3] e^{-j3}

= 1 – 2 + 1 – 2 = 2

k = 3, X_{DFT} [3] = x[n] e^{-jn3/2}

= x[10] e^{0} + x [1] e^{-j3/2} + x[2] e^{-j3} + x[3] e^{-j9/2}

= 1 – 2j – 1 + 2j = 0

so, X_{DFT} [K] = {6,0, -2, 0}

- (a)

for a band-pass signal, minimum sampling frequency is given as

f_{s} = 2fH/N

where, N = int (fH/B)

Here, f_{l} = 4 kHz, f_{h} = 6 kHz

bandwidth B = F_{H} – F_{L} = 6 – 4 = 2 kHz

N = int (6/2) = 3

f_{s} = 2 x 6/3 = 4 kHz

- (c)

x_{1} (t) * x_{2}(t) = x_{1} (f) x_{2} (f) (product of spectrum)

so, x_{1}(t)* x_{2}(t) extends only to 2 kHz

nyquist rate = 2 x 2 kHz = 4 kHz

- (a)

Because present output y(n) depends upon past output y(n – 1) and future output y(n + 1).

- (a)

for causal system output must depend upon present and past, not on future.

- (b)

The periods of individual components are

0_{1} = 2/3 2/T_{1} = 2/3 t_{1} = 3

0_{2} = 1/4 2/t_{2} = 1/4 t_{2} = 8

0_{3} = 1/3 2/t_{3} = 1/3 t_{2} = 6

0_{4} = 1/2 2/t_{4} = 1/2 t_{4} = 4

Common period is LCM of the individual periods

T = LCM (3, 8, 6, 4)

= 24